How do you find the vertex, focus and directrix of $x + y^{2} = 0$ $x+y^2=0$ ?

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Answer 1 · 2018-08-03T12:42:16

Vertex is at $(0, 0)$ $(0,0)$ , focus is at $(- 0.25, 0)$ $(-0.25,0)$ and
directrix is $x = 0.25$ $x= 0.25$

$x + y^{2} = 0 or y^{2} = - x or {(y - 0)}^{2} = - 4 \cdot \frac{1}{4} (x - 0)$ $x+y^2=0 or y^2 = -x or (y-0)^2 = -4 *1/4 (x-0)$

The equation of horizontal parabola opening left is

${(y - k)}^{2} = - 4 p (x - h); h = 0, k = 0; p = \frac{1}{4} = 0.25$ $(y-k)^2 = -4 p(x-h) ; h=0 ,k=0 ; p= 1/4=0.25$

Therefore, vertex is at $(0, 0)$ $(0,0)$ The distance between focus and

vertex is $p=0.25 :.$ Focus is at $(-0.25,0)$ . Vertex is at midway

between focus and directrix. Therefore , directrix is $x= 0.25$

graph{x+y^2=0 [-12.66, 12.65, -6.33, 6.33]} [Ans]

How do you find the vertex, focus and directrix of x+y^2=0x+y2=0x+y^2=0?